Question: Teams $ A$, $ B$, and $ C$ are playing a game of strength. Each team has attached a rope to a metal ring and is trying to pull the ring into their own area. Team $ A$ pulls with force vector ${\vec{a}} = 5\hat{i} + 2\hat{j}$. Team $ B$ pulls with force vector ${\vec{b}} = -6\hat{i} + 1\hat{j}$. Team $ C$ pulls with force vector ${\vec{c}}$ such that the ring does not move. (Forces are given in kilo newtons, $\text{kN}$.) What is the magnitude of ${\vec c}\,$ ?
Solution: Whenever forces are pulling in different directions, they tend to partially cancel each other out. This canceling effect is exactly what happens when we add vectors. In order for ${\vec c}$ to perfectly cancel out the effects of ${\vec a}$ and ${\vec b}$, the following relationship must be true: ${\vec c} + {\vec a} + {\vec b} = (0,0)$. So if ${\vec a} + {\vec b}$ is ${( 5 \hat i + 2 \hat j ) } + {( -6 \hat i + 1\hat j )} = -1 \hat i + 3 \hat j $, then ${\vec c}$ must be $1 \hat i + (-3) \hat j $. We can find the magnitude of ${\vec{c}} $ using the Pythagorean theorem. $\begin{aligned} \|{\vec{c}} \|^2 &= 1^2 + (-3)^2\\\\ \|{\vec{c}} \| &= \sqrt{1^2 + (-3)^2}\\\\ \|{\vec{c}} \| &= \sqrt{10}\\\\ \|{\vec{c}} \| &\approx 3.2 \text{ kN} \end{aligned}$ Finding the direction of ${\vec c}$ will tell us in what direction team $ C$ is pulling. ${\vec c}$ is pointing in the fourth quadrant with an $x$ -component of $1$ and a $y$ -component of $-3$. We can find the direction of any vector $\vec v$ in the fourth quadrant using the arctangent function and adding $360^\circ$. $\begin{aligned} \tan \theta &= \dfrac{ y}{ x}\\ \\ \tan \theta &= \dfrac{-3}{1}\\\\ \theta &= \arctan{\left( \dfrac{-3}{1} \right)} \\\\ \theta&\approx -72^\circ \\\\ \theta&\approx 288^\circ \end{aligned}$ The magnitude of ${\vec c}$ is $3.2 \text{ kN}$. Team $ C$ is pulling in a direction of $288^\circ$.